Ch. 20: Electrochemistry Electrochemistry is the branch of

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Ch. 20 Electrochemistry

• Electrochemistry is the branch of chemistry that
  deals with relationships between electricity and
  chemical reactions.
• We are going to study a type of reaction where
  electrons are transferred between reactants.
• These reactions are called oxidation-reduction
  reactions, or redox reactions.
• Before we begin studying redox reactions, we need
  to cover
• Oxidation Numbers
• Oxidation numbers allow us to keep track of the
  electrons gained and lost during chemical
  reactions.
• The oxidation , (or oxidation state), of an
  element in a compound is a hypothetical charge
  based on a set of rules

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Rules for Assigning Oxidation Numbers

• Rule 1 An atom in its elemental form has an
  oxidation of zero.
• Examples H2 , Ag, Br2, Pb(Ox. s zero for
  each element)
• Rule 2 Monatomic ions have an Ox. equal to
  its charge.
• Examples K (Ox. for K 1) S2- (Ox. for
  S -2)
• Rule 3 Hydrogen is 1 when attached to
  nonmetals in a compound and -1 when attached to
  metals in a hydride.
• Examples HCl (Ox. for H 1) NaH (Ox.
  for H - 1)
• ( Since the compounds are electrically
  neutral, we can then say that the Ox. of Cl
  must be -1, and Na is 1.)
• Rule 4 Fluorine is -1 in all compounds, and
  Oxygen has an Ox. of -2 in all compounds except
  peroxides when its Ox. is -1 O22-
• Examples Na2O (Ox. of oxygen -2 while Na
  is 1)
• H2O2 (Ox. of oxygen is -1 while H is
  1)
• Rule 5 The sum of the Ox. s in a compound
  is zero, and for polyatomic ions, the sum of the
  Ox. s equals the charge of the ion.
• Example HNO3(H 1, O -2, so N 5)
  SO42-(O -2, so S 6)

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Redox Reactions

• Now we can explore a redox reaction in detail
• Consider the reaction of zinc with an acid
• Zn(s) 2H(aq) ? Zn2(aq) H2(g)
• If we examine the oxidation state of zinc, we
  see that zinc started out at zero and ended up at
  2it lost 2 electrons.
• - The process in which a substance increases its
  oxidation state (by losing electrons) is called
  oxidation.
• If we examine the oxidation state of hydrogen,
  we see that hydrogen started out at 1 and ended
  up at zeroit gained an electron.
• - The process in which a substance decreases its
  oxidation state (by gaining electrons) is called
  reduction.
• Zinc reduced hydrogen, so zinc is called the
  reducing agent. (In other words, the substance
  that is oxidized is the reductant.
• Hydrogen oxidized zinc, so hydrogen is called
  the oxidizing agent. (In other words, the
  substance that is reduced is called the oxidant.

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Redox Reactions

• The easiest way to remember the difference
  between reduction and oxidation is
• LEO goes GER
• Losing Electrons Oxidation Gaining
  Electrons Reduction
• Balancing Redox Reactions
• In order to balance redox reactions we need to
  remember the following
• 1) Conservation of Mass the amount of each
  element present at the beginning of the reaction
  must be present at the end.
• 2) Conservation of Charge electrons are not
  lost in a chemical reaction. They are
  transferred from one reactant to another.
• Half-reactions are a convenient way of
  separating oxidation and reduction reactions.
• Lets look at an easy example

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Redox Reactions

• Consider the reaction
• Sn2(aq) 2Fe3(aq) ? Sn4 (aq) 2Fe2(aq)
• The oxidation half-reaction is
• Sn2(aq) ? Sn4(aq) 2e (LEO)
• (Note that electrons are shown as a productthey
  are lost.)
• The reduction half-reaction is
• 2Fe3(aq) 2e ? 2Fe2(aq) (GER)
• (Note that electrons are shown as a
  reactantthey are gained.)
• Now well look at one that is more complicated,
  but it is still simply a redox reaction.

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Balancing Redox Reactions

• Consider the titration of an acidic solution of
  Na2C2O4 with KMnO4...
• MnO4-(aq) C2O42-(aq) ? Mn2(aq) CO2(g)
• MnO4- is reduced to Mn2Manganese starts out at
  an oxidation state of 7 and is reduced to 2.
• C2O42- is oxidized to CO2Carbon starts out at an
  oxidation state of 3 and ends up at 4.
• First, we have to write the 2 half-reactions
• MnO4-(aq) ? Mn2(aq) (Reduction)
• C2O42-(aq) ? CO2(g) (Oxidation)
• Then we balance each half-reaction by following
  these steps

7
-2
3
-2
2
4
-2
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Balancing Redox Reactions
MnO4-(aq) ? Mn2(aq) C2O42-(aq) ? CO2(g) a.
First, balance the elements other than H and
O (Nothing needs to be done yet for this half
reaction.) MnO4-(aq) ? Mn2(aq)
(Balance the carbon) C2O42-(aq) ?
2CO2(g) b. Then balance O by adding
water MnO4-(aq) ? Mn2(aq) 4H2O(l)
(The oxygen is already balanced in this
half-reaction) C2O42-(aq) ? 2CO2(g)
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• (Continued) MnO4-(aq) ? Mn2(aq) 4H2O(l)
• C2O42-(aq) ? 2CO2(g)
• c. Then balance H by adding H (if in an acidic
  environment.)
• 8H MnO4-(aq) ? Mn2(aq) 4H2O(l)
• (Nothing needs to be done yet to this half
  reaction) C2O42-(aq) ? 2CO2(g)
• d. Finish by balancing charge by adding
  electrons.
• 8H MnO4-(aq) ? Mn2(aq)
  4H2O(l) (GER)
• The left side Mn is 7 while the right side Mn is
  2. We add the electrons to the left side of the
  equation(How many are gained?)
• 5e- 8H MnO4-(aq) ? Mn2(aq)
  4H2O(l)
• C2O42-(aq) ? 2CO2(g) (LEO)
• The left side is -2 and the right side is 0. We
  put the electrons on the product side which shows
  the electrons being lost.
• C2O42-(aq) ? 2CO2(g) 2e- (Each
  carbon lost one electron.)

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• (We are not done yet!!! )
• The half-reactions need to be multiplied so that
  the of electrons gained and lost is equal.
• 2 x 5e- 8H MnO4-(aq) ? Mn2(aq)
  4H2O(l)
• 5 x C2O42-(aq) ? 2CO2(g) 2e-
• Then you end up with
• 10e- 16H 2MnO4-(aq) ? 2Mn2(aq)
  8H2O(l)
• 5C2O42-(aq) ? 10CO2(g) 10e-
• Now you add the reactions together and cancel out
  items wherever possible
• 16H 2MnO4-(aq) 5C2O42-(aq) ? 2Mn2(aq)
  8H2O(l) 10CO2(g)
• It is wise to double-check and see if everything
  is balanced, including the charge!

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• What if the reaction were in a basic environment?
• Add OH- to both sides of the reaction to cancel
  out the H ions
• 16OH- 16H 2MnO4-(aq) 5C2O42-(aq)?2Mn2(aq)
  8H2O(l) 10CO2(g) 16OH-
• This results in water being formed
• 16H2O(l) 2MnO4-(aq) 5C2O42-(aq) ?
  2Mn2(aq) 8H2O(l) 10CO2(g) 16OH-
• Now we can simplify the equation by canceling out
  8H2O(l) that appears on both sides of the
  equation
• 8H2O(l) 2MnO4-(aq) 5C2O4 2-(aq)?2Mn2(aq)10CO
  2(g) 16OH-(aq)

Balancing redox reactions is one of the most
important aspects in Ch. 20! We will practice
more examples later!
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Voltaic Cells

• The energy released in a spontaneous redox
  reaction is used to perform electrical work.
• Voltaic or galvanic cells are devices in which
  electron transfer occurs via an external circuit.
• Voltaic cells are spontaneous.
• Example If a strip of Zn is placed in a
  solution of CuSO4, Cu is deposited on the Zn and
  the Zn dissolves by forming Zn2.
• Zn is spontaneously oxidized into Zn2 by the
  Cu2.
• The Cu2 is spontaneously reduced into Cu0 by
  the Zn.
• The entire process is spontaneous!

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Voltaic Cells

• In order for the redox reaction to do work, the
  half-reactions need to be separated, and the
  electrons need to be able to flow from one
  half-reaction to the other.
• Voltaic cells consist of
• Anode (where oxidation occurs) Zn(s) ? Zn2(aq)
  2e-
• Cathode (where reduction occurs) Cu2(aq) 2e-
  ? Cu(s)
• Salt bridge (used to complete the electrical
  circuit) cations move from anode to cathode,
  anions move from cathode to anode.
• The two solid metals are the electrodes
  cathode, (), and anode, ().
• Next, we will look at a picture of this voltaic
  cell and then we will analyze what is going on

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Voltaic Cells

• As oxidation occurs, Zn is converted to Zn2 and
  2e-. These electrons flow towards the cathode
  where they are used in the reduction reaction
  which turns Cu2 into Cu0.
• We expect the Zn electrode, (the anode), to lose
  mass and the Cu electrode, (the cathode), to gain
  mass.
• Electrons flow from the anode (-) to the cathode
  ().
• Electrons cannot flow through the solution they
  have to be transported through an external
  wire.
• Anions and cations move through a porous barrier
  or salt bridge.

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Voltaic Cells
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Voltaic Cells

• Anions in the salt bridge move into the anode
  compartment to neutralize the excess Zn2 ions
  formed by oxidation.
• In the same way, cations from the salt bridge
  move into the cathode compartment to neutralize
  the excess negative charge formed during
  reduction.
• Summary Anode, (-), LEO, anions move towards
  anode
• Cathode, (), GER, cations move
  towards cathode
• Electrons move from the anode to the
  cathode.
• A way to remember them ANO CPR
• anode negative
  oxidation cathode positive reduction

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Voltaic Cells
The zinc anode gets smaller and the copper
cathode gets larger. ANO LEO
CPR GER
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Voltage

• The flow of electrons from anode to cathode is
  spontaneous. What is the driving force?
• Electrons flow from anode to cathode because the
  cathode has a lower electrical potential energy
  than the anode.
• Potential difference difference in electrical
  potential.
• The potential difference is measured in volts
• One volt (V) is the potential difference
  required to impart one joule (J) of energy to a
  charge of one coulomb (C).
• VoltsJoules/Coulomb
• A Coulomb of charge is 6.25 x 1018 electrons.
• One mole of electrons is defined as having
  96,500 C of charge. This is called a faraday, F
  
• 1 F 96,500 C/mole of e

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EMF

• Electromotive force (emf) is the force required
  to push electrons through the external circuit.
• Cell potential Ecell is the emf of a cell.
• This is known as the cell voltage.
• Ecell is gt 0 for a spontaneous reaction.
• For 1 Molar solutions 1 atm pressure for
  gases, at 25º C (standard conditions), the
  standard emf (standard cell potential) is called
  Eºcell.
• Example For the zinc/copper voltaic cell
  Eºcell 1.10 V
• The emf of a cell depends on the particular
  cathode and anode half-cells that are used.

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Standard Reduction (Half-Cell) Potentials

• In order to determine the emf of a particular
  cell, you need to know the Eºred of each half
  cell involved.
• Standard reduction potentials, Eºred , are
  measured relative to a standard.
• We use the following half-reaction as our
  standard
• 2H(aq), (1 M) 2e ? H2(g), (1 atm) Eºcell 0
  V.
• This electrode is called a standard hydrogen
  electrode, (SHE).
• The SHE is assigned a standard reduction
  potential of zero.
• The Eºred for other half-reactions can be
  measured relative to the SHE.
• The emf of any cell can then be calculated from
  all of the standard reduction potentials that
  have been tabulated
• Eºcell Eºred (cathode) Eºred (anode)

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SHE and Zinc Anode
Zinc is oxidized. Since the cells emf is 0.76
V, that means the Eºred for Zn2(aq) 2e-
?Zn(s) would be 0.76 V.
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Eºcell

• Practice Problem The standard emf for the
  following cell is 1.46 V.
• In(aq) Br2(l) ? In3 (aq) Br-(aq)
• Calculate the Eºred for the reduction of In3 to
  In.
• First, lets break this up into the two
  half-reactions
• (LEO ANO) In(aq) ? In3(aq) 2e-
  Eºred ?
• (GER CPR) 2e- Br2(l) ? 2Br-(aq)
  Eºred 1.06 V
• Plug and Chug E ºcell Eºred (cathode) - Eºred
  (anode)
• 1.46 1.06 Eºred (anode)
• Therefore Eºred (anode) 0.40 V
• Note This reaction is spontaneous. Also, In
  was oxidized in this reaction, so you would
  expect the value for Eºred to be (-).

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Eºcell

• Changing the coefficients of a half-reaction
  does not change Eºred.
• Zn2(aq), (1 Molar) ? Zn(s) Eºred
  0.76 V
• 2Zn2(aq), (1 Molar) ? 2Zn(s) Eºred
  0.76 V
• As you may suspect, the concentration will
  change Eºcell as we will see later on in the
  notes.
• The cathode reaction (CPR) of a voltaic cell
  will always have the more positive Eºred than the
  reaction at the anode (ANO).
• In essence, the greater driving force of the
  cathode half-reaction is used to force the anode
  reaction to occur in reverse as an oxidation.
• The entire process ends up being spontaneous.

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Eºcell

• Practice Problem A voltaic cell is based on the
  following two standard half-reactions
• Cd2(aq) 2e- ? Cd(s)
• Sn2 (aq) 2e- ? Sn(s)
• Which reaction takes place at the cathode and
  which is at the anode?
• Look up the Eºcell for each in Appendix E
• Cd2 0.403 V Sn2 0.136 V
• Since Sn2 is more positive, then it will be
  reduced at the cathode.
• Therefore, Cd2 will be oxidized at the anode.
• What is the value for Eºcell?
• Eºcell Eºred (cathode) Eºred (anode)
• Eºcell 0.136 V ( 0.403) 0.267 Volts
• Note The reaction had to be positive (and
  spontaneous) for a voltaic cell!!

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Oxidizing and Reducing Agents

• We can use this table to determine the relative
  strengths of reducing (and oxidizing) agents.
• The more positive E?red the stronger the
  oxidizing agent on the left.
• The more negative E?red the stronger the
  reducing agent on the right.
• We can use this to predict if one reactant can
  spontaneously oxidize another.

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Oxidizing and Reducing Agents
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EMF and Free Energy

• How does emf relate to ?G?
• ?G nFE
• where... n of electrons transferred during
  the process
• F 96,500 C/mole
• E cell potential
• Since n and F are positive, if E gt 0 then ?G lt 0
  and the reaction would be spontaneous.
• If the reactants and products are in their
  standard states, then
• ?Gº nFEº

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Effect of Concentration on Cell EMF

• The Nernst Equation
• A voltaic cell will function until E 0, at
  which point equilibrium has been reached.
• The point at which E 0 is determined by the
  concentrations of the species involved in the
  redox reaction.
• The Nernst equation relates emf to concentration
  
• RT lnQ
• or expressed as log base 10
  2.303 RT
• (See p.799 in the text for the derivation.)

E Eº
nF
log Q
E Eº
nF
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The Nernst Equation

• When T 25º C, the equation is simplified to
• 0.0592 V
• Lets see how this equation can be used
• Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
• At standard conditions, the Zinc/Copper cell
  has an emf of 1.10 V. What is the voltage of
  the cell at 25º C when Cu2 5.0 M and
  Zn2 .050 M?
• Recall that Q products/reactants.
  (Remember, pure solids and liquids do not appear
  in the expression for Q!)
• Therefore E 1.10 V 0.0592 V log
  0.050/5.0
• Since Q lt 1, then log Q lt 0, and this will
  increase the voltage
• E 1.16 Volts

E Eº
log Q
n
2 e
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Concentration Cells

• The Nernst equation shows us that using the same
  species in the anode and cathode compartments of
  a cell but at different concentrations then a
  voltage will be generated.
• For example
• If the concentrations of Ni2 are unequal, a
  voltage will be generated.

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Concentration Cells

• At the anode, Ni(s) ? Ni2 increasing the
  concentration of nickel ions in this compartment.
• At the cathode, Ni2 ? Ni(s) and the
  concentration of nickel ions in this compartment
  is reduced.
• Therefore
• Q Ni2dilute/Ni2concentrated
• At 25º C
• E 0 0.0592 log 0.001/1 0.0888 V
• When the compartments reach equilibrium, emf
  zero.
• The pH meter functions on this principle, and so
  does your heartbeat!

2 e-
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Cell EMF and Chemical Equilibrium

• A voltaic cell is functional until E 0 at
  which point equilibrium has been reached. The
  cell is then dead.
• A system is at equilibrium when ?G 0.
• From the Nernst equation, at equilibrium and 298
  K (E 0 Volts and Q Keq)
• 0 Eº RT ln Keq
• This equation can be simplified at 25º C to
• log Keq nEº/0.0592
• So you can calculate Keq of a redox reaction by
  knowing Eº or vice versa!

nF
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Batteries

• A battery is a portable, self-contained
  electrochemical power source consisting of one or
  more voltaic cells.
• The () end of the battery is called the
  cathode, and the () end is called the anode.
• If you wanted to generate 9 V, you will have to
  link together multiple cells in series since no
  single redox reaction can produce that much
  voltage.
• Primary cells cannot be recharged.
• Secondary cells are rechargeable.

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Types of Batteries

• Lead-Acid Battery (12 V car battery)
  
• Cathode lead dioxide plate
• PbO2(s) HSO4-(aq) 3H(aq) 2e- ? PbSO4(s)
  2H2O
• Anode (lead plate)
• Pb(s) HSO4-(aq) ? PbSO4(s) H(aq) 2e-
• The overall electrochemical reaction is
• PbO2(s) Pb(s) 2SO42-(aq) 4H(aq) ?
  2PbSO4(s) 2H2O(l)
• E?cell E?red (cathode) E?red (anode)
• E?cell (1.685 V) ( 0.356 V)
• E?cell 2.041 V.
• Six of these cells are linked in series to make
  12 V, wood or glass-fiber spacers are used to
  prevent the electrodes from touching.

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Lead Acid Batteries

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Lead-Acid Batteries

• Lead-Acid batteries can be recharged.
• H2SO4 decreases as the battery discharges and
  increases as the cell is charged back up while
  the reactions run in reverse.
• H2SO4 is much more dense than H2O, so the
  condition of the battery can be monitored by
  testing the specific gravity, (density), of the
  solution
• specific gravity 1.25 to 1.30 charged up and
  ready to go!
• specific gravity lt 1.20 needs charging
• These batteries are called wet cells because
  the electrolyte is a liquid.

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Alkaline Batteries (Dry Cells)

• The most common nonrechargeable battery is the
  alkaline battery.
• Powdered zinc metal is immobilized in a gel in
  contact with a concentrated solution of KOH.
• The electrolyte is a base, thus these batteries
  are alkaline.
• The electrolyte is a paste, hence the name dry
  cell.
• The reaction at the anode is
• Zn(s) 2OH-(aq)? Zn(OH)2(s) 2e-
• The reaction at the cathode is the reduction of
  MnO2
• 2MnO2(s) 2H2O(l) 2e- ? 2MnO(OH)(s) 2OH-(aq)
• The cell potential of these batteries is 1.55 V
  at room temperature.

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Alkaline Batteries (Dry Cells)

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Other Batteries

• A common rechargeable battery is the
  nickelcadmium (NiCad) battery.
• The reaction at the cathode is
• 2NiO(OH)(s) 2H2O(l) 2e- ? 2Ni(OH)2(s)
  2OH-(aq)
• The reaction at the anode is
• Cd(s) 2OH-(aq) ? Cd(OH)2(s) 2e-
• The cell potential of this battery is about 1.30
  V at room temp.
• Cadmium is a toxic heavy metal.
• There are environmental concerns to be addressed
  with respect to disposal of such batteries.
• Other rechargeable batteries have been
  developed.
• - NiMH batteries (nickelmetalhydride).
• - Liion batteries (lithiumion
  batteries)(greater energy density)

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Corrosion

• An example of an undesirable redox reaction is
  the corrosion of metals.
• A metal is attacked by a substance in the
  environment and converted to an unwanted
  compound.
• Consider the corrosion (rusting) of iron
• Since Eºred(O2) gt Eºred(Fe2), iron can be
  oxidized by O2.
• Cathode
• O2(g) 4H(aq) 4e- ? 2H2O(l) Eºred
  1.23 V.
• Anode
• Fe(s) ? Fe2(aq) 2e- Eºred -0.44 V.

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Corrosion

Dissolved oxygen in water usually causes the
oxidation of iron. The Fe2 initially formed
can be further oxidized to Fe3, which forms
rust, Fe2O3xH2O(s). Oxidation occurs at the
site with the greatest concentration of O2.
Other factors to consider are the pH, presence of
salts, stress on the iron, and contact with other
metals.
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Corrosion of Iron

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Preventing the Corrosion of Iron

• Corrosion can be prevented by coating the iron
  with paint or another metal to keep it away from
  O2 and water.
• Another way to protect the iron is by a process
  called cathodic protection.
• Coating iron with a layer of zinc, galvanized
  iron, can protect the iron from corrosion even
  after the surface coat is broken.
• Zn2(aq) 2e- ? Zn(s) Eºred - 0.76 V
• Fe2(aq) 2e- ? Fe(s) Eºred - 0.44 V
• The standard reduction potentials indicate that
  Zinc is easier to oxidize than iron, so the Zinc
  will be oxidized!
• Zinc is the sacrificial anode and is slowly
  destroyed. Iron acts as the cathode where the
  oxygen is reduced.
• We can use something similar to protect
  underground pipelines.
• Often, Mg is used as a sacrificial anode(Eºred
  - 2.37 V)

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Preventing the Corrosion of Iron

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Preventing the Corrosion of Underground Pipes

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Electrolysis

• It is possible to cause a nonspontaneous redox
  reaction to occur. Such processes are driven by
  an outside source of electrical energy.
• These reactions are called electrolysis
  reactions and take place in electrolytic
  cells.
• In voltaic and electrolytic cells, reduction
  occurs at the cathode, and oxidation occurs at
  the anode.
• However, in electrolytic cells, electrons are
  forced to flow from the anode to the cathode.
• In electrolytic cells the anode is positive and
  the cathode is negative.
• (In voltaic cells the anode is negative and the
  cathode is positive.)
• Heres what an electrolytic cell looks like

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Electrolytic Cell

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Electrolysis of Molten NaCl

• Example of Electrolysis
• Cathode (GER) 2Na(l) 2e- ? 2Na(l)
• Anode (LEO) 2Cl-(l) ? Cl2(g) 2e-
• A battery or some other source of direct current
  acts as an electrical pump pushing electrons to
  the cathode and pulling them from the anode.
• As Na(s) is produced at the cathode () ,
  additional Na migrates in.
• As Cl2(g) is produced at the anode (),
  additional Cl- migrates in.
• Industrially, electrolysis is used to produce
  metals like aluminum and gases like chlorine.

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Electrolysis of Aqueous Solutions

• Do we get the same products if we electrolyze an
  aqueous solution of the salt?
• No! Water complicates the issue.
• Example Consider the electrolysis of NaF(aq)
• Cathode (GER) Na(aq) e- ? Na(s) Eºred -
  2.71 V
• Cathode (GER) 2H2O(l) 2e- ? H2(g)
  2OH-(aq) Eºred - 0.83 V
• Thus water is more easily reduced than the
  sodium ion!
• Anode (LEO) F-(aq) ? F2(g) 2e- Eºred
  2.87 V
• Anode (LEO) 2H2O(l) ? O2(g) 4H(aq)
  4e- Eºred 1.23 V
• Thus it is easier to oxidize water than the
  fluoride ion.
• What you end up doing is electrolyzing water
  into H2(g) and O2(g)!

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Electrolysis with Active Electrodes

• Active electrodes electrodes that take part in
  electrolysis.
• Example of active electrodes electroplating
• Consider an active Ni electrode and another
  metallic electrode (steel) placed in an aqueous
  solution of NiSO4
• Anode Ni(s) ? Ni2(aq) 2e- Eºred
  0.28 V
• Cathode Ni2(aq) 2e- ? Ni(s)
  Eºred 0.28 V
• Ni gets transferred from the anode to the
  cathode and the Ni plates on the inert (steel)
  cathode.
• (Of course a small emf will be needed to push
  the reaction along.
• Electroplating is important in protecting
  objects from corrosion
• Example Stainless steel

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Electrolysis with Active Electrodes

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Quantitative Aspects of Electrolysis

• We want to know how much material we obtain with
  electrolysis.
• Consider the reduction of Cu2 to Cu
• Cu2(aq) 2e- ? Cu(s)
• 2 moles of electrons will plate 1 mole of Cu.
• Using this information as well as the following
  conversion factors, we can calculate how much
  copper can be obtained from electrolysis given a
  certain amount of current
• 1 mole of electrons 96,500 Coulombs 1 faraday
• 1 Ampere of current 1 Coulomb/second (1A
  1C/s)

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Quantitative Aspects of Electrolysis

• Practice Problem How much copper (in grams) can
  be obtained from electrolysis by using 3.5 amps
  for 5 hours?
• First, determine the amount of coulombs

3600 s
5 hr
3.5C
63,000 C
x

x
1 hr
1 sec

• Then convert coulombs to grams of copper

1mole e-
1 mole Cu
63.5 grams
20.7 g of Cu
x

63,000 C
x
x
96,500 C
2 moles e-
1 mole Cu